Dirichlet?

Calculus Level 3

Does this series converge or diverge?

1 3 + 1 5 + 1 7 1 9 1 11 + 1 13 + 1 15 + 1 17 1 19 1 21 + \displaystyle \frac{1}{3} + \frac{1}{5} + \frac{1}{7} - \frac{1}{9} - \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \frac{1}{17} - \frac{1}{19} - \frac{1}{21} + \ldots

It converges It diverges It is unsolved Can't be found

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1 solution

Chris Lewis
Feb 10, 2020

Let S N S_N be the sum of the first N N terms of the series in question. We'll show it diverges with the comparison test.

Note that the signs of the terms repeat every five terms. If we concentrate on the first five terms, we just have to observe that 1 3 > 1 9 \blue{\frac13>\frac19} and 1 5 > 1 11 \green{\frac15>\frac{1}{11}} :

S 5 = 1 3 + 1 5 + 1 7 1 9 1 11 > 1 7 S_5=\blue{\frac13}+\green{\frac15}+\frac17-\blue{\frac19}-\green{\frac{1}{11}}>\frac17

We can do this for every set of five terms; so

S 5 k > 1 7 + 1 17 + 1 27 + + 1 10 k 3 S_{5k}>\frac{1}{7}+\frac{1}{17}+\frac{1}{27}+\cdots+\frac{1}{10k-3}

Since 1 10 k 3 > 1 10 k \frac{1}{10k-3}>\frac{1}{10k} , we have

S 5 k > 1 10 + 1 20 + 1 30 + + 1 10 k = 1 10 H k S_{5k}>\frac{1}{10}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{10k}=\frac{1}{10} H_k

where H k H_k is the k t h k^{th} harmonic number ; the harmonic series diverges and hence so does the series in question.

That was my mental argument. Thanks for making it concrete for me!

Steven Perkins - 1 year, 4 months ago

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